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In an L-R-C series circuit, L = 0.280 H and C = 4.00 mF. The voltage amplitude of the source is 120 V. (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 1.70 A. What is the resistance R of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Short Answer

Expert verified

a)The resonance angular frequency of the circuit is 945rads.b) Resistance R of the resistor 70.6Ω.c)The peak voltages across the inductor is 450 V, the capacitor is 450V and the resistor is 120V.

Step by step solution

01

Step-1: Formulas used  

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

Z=R2+(XL-XC))2, where Z is the impedance

The equivalent Ohm’s law relation to get the amplitude voltage V in the circuit.

V=IZ

Similarly, the amplitude voltage across the resistor, capacitor and inductor is found by the relation

V=IX

Where X is the reactance which is equal toXL=Ó¬³¢ for inductor andXC=1Ó¬°ä for capacitor and Rfor resistor.

The process of peaking the current at a particular frequency represents the resonance and that frequency is known as resonant frequency.

Ó¬=1LC,whereis the resonant angular frequency.

At resonance, XL=XC.

At resonance, the impedance is minimum and this minimum value is same as that of resistor.

The maximum voltages across R, C and L is

VR=IRVC=IXCVL=IXL

02

Step-2: Calculations for resonant angular frequency.

Ó¬=1LCL=0.280mHC=4.00μ¹ó

Ӭ=10.280H4×10-6C=945rads

Therefore, the resonance angular frequency of the circuit is 945rads.

03

Step-3: Calculations for resistance.

At resonance,

Z=R

Z=VI

I=1.70AV=120VZ=120V1.70A=70.6ΩZ=RR=70.6Ω

Therefore, Resistance R of the resistor .70.6Ω

04

Step-4: Calculations for voltage amplitude across resistor, capacitor, inductor.

VR=120VVC=1Ó¬°ä=1.70A(945rads)4×10-6F=450VVL=IÓ¬³¢=1.70A(945rads)0.280H=450V

At resonance,VR is the same as the sourceVL andVC are the same, andVL-VC =0.

Therefore, a)The resonance angular frequency of the circuit is 945rads.b) Resistance R of the resistor 70.6Ω.c)The peak voltages across the inductor is 450V, the capacitor is and the resistor is 120V.

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